As above, I'm trying to find all linear operators $L: C([0,1]) \to C([0,1])$ which satisfy the following 2 conditions:

I) $Lf \, \geq \, 0$ for all non-negative $f\in C([0,1])$.

II) $Lf = f$ for $f(x)= 1$, $f(x)=x$, and $f(x)=x^2$.

I'm honestly not sure where to start here - I'm struggling to use these conditions to pare down the class of linear operators which could satisfy the conditions significantly. Could anyone help me get a result out of this? Thank you!

Its proof (as well as the argument above) is a variant of the usual proof of the Weierstrass approximation theorem using Bernstein polynomials. One may take
\[
L_{n}f(x) = \sum_{k = 0}^{n} \begin{pmatrix} n \\ k \end{pmatrix}x^{k}(1-x)^{n-k} f(k/n),
\]
in Korovkin's theorem and verify directly that $L_n g_{i} \to g_{i}$ for $i = 0,1,2$, so Korovkin's theorem yields the Weierstrass approximation theorem.

I'm sorry, I don't think i'm quite there yet - I can see how we can use your second point to get $|Lf(x)-Lf(y)|<\epsilon+C(x-y)^2$, but what can we make use of this bound for? If we treat y as constant we can say $|Lf(x)-f(y)|<\epsilon+C(x-y)^2$ too but I'm not sure how that helps. As we take a smaller epsilon, our C will become arbitrarily large, so presumably we can't really guarantee much accuracy here? Sorry about this...
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Stephen GreggJan 8 '11 at 15:39

You're almost there. Once you've fixed $\varepsilon$ and $\delta$, the inequality in the second point holds for all $x,y \in [0,1]$! To prove this you need to to consider two cases: $|x - y| < \sqrt{\delta}$ and $|x - y| \geq \sqrt{\delta}$.
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t.b.Jan 8 '11 at 15:48

Oh, yes, I just got there and then saw your post! In fact, can't we just evaluate my second inequality in the previous post at x=y? At which point we'd get $|Lf(y)-f(y)|< \epsilon$?
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Stephen GreggJan 8 '11 at 15:53

Yes, of course, but you need to know that the inequality holds for all $x \in [0,1]$ (so that it is an inequality of continuous functions on $[0,1]$ and you can use the positivity of $L$).
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t.b.Jan 8 '11 at 15:55

1

@Stephen: You're welcome, that was fun! Could you accept one of my posts as answer, please? Then the question will stop showing up as unanswered.
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t.b.Jan 8 '11 at 23:55